In the **early afternoon** or high in the sky around the time, the **sun** is setting.

The** crescent moon** rises around noon and sets around midnight. It can be seen near the horizon in the evening or high in the sky as the sun sets. If you live in the northern hemisphere, the first quarter moon appears as the right half of the illuminated surface. **Occurs** when the moon is at right angles to the sun as seen from Earth.

The right half of the **moon** appears bright and the left half appears dark. During the period from the new moon to the first quarter moon, the **illuminated **part of the moon grows larger day by day and continues to grow until the full moon. The first quarter is also called the crescent moon because 50% of its surface is briefly illuminated by the sun. Usually, he takes 3 nights. Each phase of the moon has a profound effect on humans and the earth.

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Sound _____. does not need a medium to travel is reflected when it bounces off a shiny surface moves in longitudinal waves is created by electric and magnetic fields

**Answer:**

reflection

**Explanation:**

it is a regular reflaction of light

Element x is a non metal in which position of the periodic table can element x be found

**Answer:**

To the right

**Explanation:**

Metals are to the left, nonmetals are to the right. There's a border I drew that separates nonmetals and metals

_______ is the study of the way all living things relate to each other in the world. The word comes from two Greek words which mean the study or science of the home.

**Ecology** is the study of the way all **living** things relate to each other in the world.

This is referred to as the science which deals with the study of **organisms** and how they interact with their **environment.** It comprises of the study of the organisms present and their habitat.

Their **habitat** is the area where they live which comprises of many **resources** such as food, water etc which are needed for their growth and survival. The **living** things interact with it for beneficial reasons and is the reason why **Ecology** was chosen as the correct choice.

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The object represented by line E does not stop. Is this true or false?

The given graph is about velocity and time, the dependent and independent variable, respectively. As we can observe, line E crosses the zero level to the negative zone, which means the object didn't stop but changed its direction.

Therefore, the statement is true.WHAT IS CHROMATOGRAPHY ?

Answer:

Chromatography is a process for separating components of a mixture. To get the process started, the mixture is dissolved in a substance called the mobile phase, which carries it through a second substance called the stationary phase.

Explanation:

hope it will be helpful

**Answer:**

**Chromatography is a process for separating components of a mixture. **

**Explanation:**

**To get the process started, the mixture is dissolved in a substance called the mobile phase, which carries it through a second substance called the stationary phase.**

**Have a good day!**

Part 1 Assume that as a wave travels from one particular medium to another, its speed decreases. What happens to the wave's frequency?A. Its frequency increases.B. Its frequency decreases.C. Its frequency remains the same.D. It is impossible to determine without more data.Part 2Which best explains the correct answer to Part 1?A. A wave's frequency always remains the same. B. A wave's frequency is inversely proportional to its speed.C. Given that the wave's wavelength remains the same, if the wave's speed decreases, then the frequency must decrease.D. Given that a wave's speed is equal to the product of its wavelength and frequency, the wavelength must be known to determine the frequency.

Answer:

Part 1

C. Its frequency remains the same.

Part 2

A. A wave's frequency always remains the same.

Explanation:The relationship between wavelength, frequency, and speed is given as:

[tex]v=\lambda f[/tex]When the frequency of a wave increases, its wavelength also increases.

The **speed of a wave does not change as its frequency changes**, and vice-versa.

Therefore, **as the speed of the wave decreases, its frequency remains the same**

A wave's frequency always remains the same regardless of the speed and wavelength

9.A stock car is moving at 25.0 m/s when the driverapplies the brakes. If it stops in 3.00 s, what is itsaverage acceleration?

The average acceleration **a** of an object that changes its speed from **v_0** to **v_f** during a time interval **t** is:

Since the car is initially moving at 25.0m/s and it stops in 3.00 seconds, then its final speed is 0m/s.

Replace **v_f=0, v_0=25.0m/s** and **t=3.00s** to find the average acceleration of the car:

**Therefore, the average acceleration of the car when it stops from 25.0m/s in 3.00 seconds, is ****-8.33m/s^2**.

your pet hamster sits on a record player that has constant angular speed . if the hamster moves to a point twice as far from the center, then its linear speed

Força leonina raios solares

The table below shows part of the operating budget of a small dairy farm for the last year. The only expense not listed is maintenance. LAST YEARS OPERATING BUDGETExpense Fraction of BudgetFood 1/3Housing 1/3 Medical Care 1/4This year the managers of the farm will change the fraction for the budget for housing to ⅛ but will leave the fraction of the budget for good and medical care the same. Again, the remaining portion of the budget will be before maintenance expenses. What is the difference between the fraction of the budget for maintenance this year and last year?Show your workAnswer

[tex]\frac{5}{24}[/tex]

**Explanation**

**Step 1**

let the fractions

[tex]\begin{gathered} Food\text{ }\frac{\text{1}}{3} \\ \text{ Housing }\frac{\text{1}}{3} \\ \text{Medical care }\frac{1}{4} \\ \text{ maintenance= M ( unknown)} \end{gathered}[/tex]so

a)This year the managers of the farm will change the fraction for the budget for housing to 1/8

so

[tex]\begin{gathered} Food\text{ }\frac{\text{1}}{3} \\ \text{ Housing }\frac{\text{1}}{8} \\ \text{Medical care }\frac{1}{4} \\ \text{ maintenance= M ( unknown)} \end{gathered}[/tex]b) hence,the total budget is

[tex]\begin{gathered} \text{past year= Operatingbudget= food+housidn +Medical care+maintenance} \\ \text{replace} \\ past\text{ year= }\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+m \\ past\text{ year=}\frac{11}{12}+m \end{gathered}[/tex]this year

[tex]\begin{gathered} \text{past year= Operatingbudget= food+housidn +Medical care+maintenance} \\ \text{replace} \\ thisyear\text{= }\frac{1}{3}+\frac{1}{8}+\frac{1}{4}+m \\ thisyear\text{=}\frac{17}{24}+m \end{gathered}[/tex]**Step 2**

find the fractions for maintenance on each year:

as we are using fraction of the total budget, we need to set the sum of the fractions equals 1, hence

a) past year

[tex]\begin{gathered} \text{total = 1, so} \\ past\text{ year=}\frac{11}{12}+m=1 \\ m=1-\frac{11}{12} \\ m=\text{ }\frac{1}{12} \end{gathered}[/tex]b) this year

[tex]\begin{gathered} \text{total = 1, so} \\ this\text{year=}\frac{17}{24}+m=1 \\ m=1-\frac{17}{24} \\ m=\text{ }\frac{7}{24} \end{gathered}[/tex]**Step 3**

now, find the difference

[tex]\begin{gathered} \text{difference =}\frac{7}{24}-\frac{1}{12} \\ \text{difference =}\frac{5}{24} \end{gathered}[/tex]therefore, **the answer is 5/24**

I hope this helps you

In the experiment, the pressure of the gas is 1.2 x10^5 Pa at a temperature of 25.0°C.

When the cylinder is heated, the pressure reaches 2.1x10 Pa. Calculate the

temperature of the gas (in "C) at this pressure.

The **temperature **of the gas (in °C) at this **pressure **is 248.5 °C

**Temperature **is a bodily quantity that expresses the hotness of count or radiation. There are 3 kinds of temperature scales. Temperature is the degree of **hotness **or **coldness **of an item.

**Temperature **is a degree of the common kinetic energy of the debris in an object. whilst the temperature increases, the motion of those particles also increases. Temperature is measured with a **thermometer or calorimeter**.

Given;

P₁ = 1.2 x10⁵ Pa = 1.18430792

T₁ = 25.0°C = 298 K

P₂ = 2.1x10 Pa = 0.000207253886

T₂ = ?

using** ideal gas equaion:-**

PV = nRT

P₁/T₁ = P₂/T₂

T₂ = P₂T₁ /P₁

= 2.1x10⁵ x 298 / 1.2 x10⁵

= 521.5 K

= 248.5 °C

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1) What is the horizontal force on block A due to block B?2) What is the net horizontal force on block B?3) What is the horizontal force on block B due to block C?

Given data:

Mass of each block;

[tex]m=12\text{ kg}[/tex]Acceleration;

[tex]a=1.2\text{ m/s}^2[/tex]The free-body diagram for A,

The free-body equation for A is given as,

[tex]F-N_1=ma\ldots(1)[/tex]The free-body diagram for B is given as,

The free-body equation for B is given as,

[tex]\begin{gathered} F_B=N_1-N_2 \\ ma=N_1-N_2\ldots(2) \end{gathered}[/tex]The free-body equation for C is given as,

The free-body equation for C is given as,

[tex]\begin{gathered} F_c=ma \\ N_2=ma\ldots(3) \end{gathered}[/tex]Equating equation (2) and (3),

[tex]\begin{gathered} N_1-N_2=N_2_{} \\ N_1=2N_2 \\ N_1=2ma \end{gathered}[/tex]Part (1),

The horizontal force on block A due to block B is given as,

[tex]\begin{gathered} F_{AB}=N_1 \\ =2ma \end{gathered}[/tex]Substituting all known values,

[tex]\begin{gathered} F_{AB}=2\times(12\text{ kg})\times(1.2\text{ m/s}^2) \\ =28.8\text{ N} \end{gathered}[/tex]Therefore, **the net horizontal force on block A due to block B is 28.8 N.**

Part (2)

The net horizontal force on block B is given as,

[tex]\begin{gathered} F_B=N_1-N_2 \\ =2ma-ma \\ =ma \end{gathered}[/tex]Substituting all known values,

[tex]\begin{gathered} F_B=(12\text{ kg})\times(1.2\text{ m/s}^2) \\ =14.4\text{ N} \end{gathered}[/tex]Therefore,** the net horizontal force on block B is 14.4 N.**

Part (3)

The horizontal force on block B due to block C is given as,

[tex]\begin{gathered} F_{BC}=N_2 \\ =ma \end{gathered}[/tex]Substituting all known values,

[tex]\begin{gathered} F_{BC}=(12\text{ kg})\times(1.2\text{ m/s}^2) \\ =14.4\text{ N} \end{gathered}[/tex]Therefore, **the horizontal force on block B due to block C is 14.4 N.**

Set the cannon to have an initial speed of 25 m/s at an angle of 30-degrees. Find how long it takes to hit the ground. Will the time it takes to get to the highest point be less than ½ the time, more than ½ the time, or ½ the time?

Question 7 options:

1/2 the time

more than 1/2 the time

less than 1/2 the time

The time taken by the **projectile **to reach the **highest **point would be one half of the total **time **of the projectile , therefore the correct answer is option D .

It is the motion of any object or body when it is **ejected **off the surface of the earth and follows any curving path while being affected by the **gravitational **pull of the planet .

As given in the problem Set the cannon to have an **initial **speed of 25 m / s at an **angle **of 30 degrees .

Thus, the right response is option D because it would take the **projectile **half as long to reach the **highest **position.

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Exit Slip!

Explain how Earth's tilted

axis and its revolution

around the Sun produces

seasons.

Earth's tilted **axis **and **revolution** around the sun causes the seasons.

Different parts of **Earth** receive the Sun's direct rays, throughout the year. So, it's summer in the Northern Hemisphere when the North Pole tilts toward the **Sun**. It's winter in the Northern Hemisphere when south Pole tilts toward the Sun.

As we know that the **Earth** rotates with a constant speed, so that every hour the direct beam will traverse across a single standard meridian. **Rotation** of Earth 15 degrees is equivalent to unit of one hour . When Earth rotates in such a way that the beam of the sun shifts +1∘ of longitude from East to West and time taken is 4 minutes.

The tilt of the Earth's axis is important because it governs the strength of warming of the sun's energy. The** tilt **of the surface of the Earth also causes light to be spread across a larger area of land which is termed as called the cosine **projection effect**.

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A solid cylinder (mass 0.274 kg, radius 2.00 cm) rolls without slipping at a speed of 5.00 cm/s. What is its total kinetic energy?

**Given data**

*The given mass of the solid cylinder is **m** = 0.274 kg

*The given radius of the cylinder is **r** = 2.00 cm = 0.02 m

*The given speed is **v **= 5.00 cm/s = 0.05 m/s

The formula for the total kinetic energy is given as

[tex]\begin{gathered} U_T=U_k+U_R \\ U_T=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \\ =\frac{1}{2}mv^2+\frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 \end{gathered}[/tex]*Here **U_K** is the translation kinetic energy

*Here **U_R** is the rotational kinetic energy

*Here '**I'** is the moment of inertia of the solid cylinder

Substitute the known values in the above expression as

[tex]\begin{gathered} U_T=\frac{1}{2}(0.274)(0.05)^2+\frac{1}{2}(\frac{1}{2}\times0.274\times(0.02)^2)(\frac{0.05}{0.02})^2 \\ =0.000342+0.000171 \\ =5.13\times10^{-4}\text{ J} \\ =5.13\times10^{-1}\text{ mJ} \end{gathered}[/tex]Hence, the total kinetic energy is** U_T = 5.13 × 10^-1 mJ**

If the displacement, velocity and acceleration at an instant of a particledescribing S.H.M are respectively 7.5m, 7.5m/s, 7.5m/s?. Calculate themaximum velocity of the particle.

We are given the displacement, velocity, and acceleration of a particle that describes Simple Harmonic Motion. We are asked to determine the maximum velocity of the particle. To do that we can use the following equation for the maximum velocity of a particle describing SHM:

[tex]v_{máx}=-A\omega[/tex]Where:

[tex]\begin{gathered} A=\text{ amplitude} \\ \omega=\text{ angular frequency} \end{gathered}[/tex]To determine the values of amplitude and angular frequency we can use the expressions for displacement, velocity, and acceleration. The expression for displacement is:

[tex]x=A\cos (\omega t)[/tex]The expression for the velocity is:

[tex]v=-A\omega\sin (\omega t)[/tex]And the expression for acceleration is:

[tex]a=-A\omega^2\cos (\omega t)[/tex]Now, from the expression for the displacement we can solve for the amplitude, like this:

[tex]\frac{x}{\cos(\omega t)}=A[/tex]Now we can replace this in the expression got eh acceleration:

[tex]a=-\frac{x}{\cos(\omega t)}\omega^2\cos (\omega t)[/tex]Simplifying:

[tex]a=-x\omega^2[/tex]Now we can solve for the angular frequency:

[tex]-\frac{a}{x}=\omega^2[/tex]Taking square root to both sides:

[tex]\sqrt[]{-\frac{a}{x}}=\omega[/tex]replacing the values:

[tex]\sqrt[]{-\frac{-7.5\frac{m}{s^2}}{7.5m}}=\omega[/tex]Solving the operations:

[tex]1s^{-1}=\omega[/tex]Now we divide the formula for the displacement and the formula for the velocity, we get:

[tex]\frac{v}{x}=\frac{-A\omega\sin (\omega t)}{A\cos (\omega t)}[/tex]Simplifying we get:

[tex]\frac{v}{x}=-\omega\tan (\omega t)[/tex]Replacing the known values:

[tex]\frac{-7.5\frac{m}{s}}{7.5m}=-(1s^{-1})\tan (t)[/tex]Simplifying:

[tex]-1=-\tan (t)[/tex]Solving for "t":

[tex]t=\arctan (1)=0.78[/tex]Now we can replace these values in the formula for displacement to get the value of the amplitude:

[tex]x=A\cos (\omega t)[/tex]Solving for the amplitude:

[tex]\frac{x}{\cos (\omega t)}=A[/tex]Replacing the known values:

[tex]\frac{7.5m}{\cos (0.78)}=A[/tex]Solving the operations:

[tex]10.6m=A[/tex]Now we replace the values in the formula for the maximum velocity:

[tex]v_{\text{max}}=A\omega[/tex]Replacing:

[tex]\begin{gathered} v_{\max }=(10.6m)(1s^{-1}) \\ v_{\max }=10.6\text{ m/s} \end{gathered}[/tex]Therefore, the maximum velocity is 10.6 meters per second.

Consider a 1500kg rollercoaster moving at 4.5m/s at the top of a

30.0m hill. If instead, friction does 52kJ of work on the rollercoaster on its

trip down the hill, how fast will the rollercoaster be moving once it

reaches ground level?

The rollercoaster be moving once it reaches** ground level **with a **speed **of 23.21 m/s.

Applying **work-energy theorem**,

Work done by **conservative forces **on the body + work done by** none conservative** forces on the body = Total **change in Kinetic Energy** of the body.

Here, conservative forces are gravitational force and non-conservative force is frictional force,

MgH - Fr = 1/2MV² - 1/2MU²

Where,

M is mass of roller coaster,

g is **acceleration due to gravity**,

H is the height of the roller coaster,

Fr is the** frictional force,**

V is the **final velocity** of roller coaster,

U is the **initial velocity **of the roller coaster,

1500(9.8)(30) - 52000 = 1/2(1500)(V²-20.25)

(441000 -52000)2/1500 = V² - 20.25

518.67 = V² - 20.25

V² = 538.9

V = 23.21 m/s.

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A projectile is fired horizontally at an angle of 70° with the speed of 0.11km/s. Find the range and height. The final answer should be in meters.

In the given question **projectile** is fired horizontally at an angle of **70**° with the **speed** of **0.11km/s **,(=110m/s).

to find the **range** and **height**

**height**=544.66m

time= 10.54 second

**range**=793.63m

vy=0

h=v^2sin^2θ/ 2g

v=0.11km/s =110m/s

θ=70°

**range**=v^2sin2θ/ g

horizontal velocity,

vx= 110cos70=37.622 m/s

vertical velocity,

vy=110sin70=103.36 m/s

using the equation of motion

s=ut+1/2 at^2

**range**, s=0, t=total time taken

**range**= 110^2 sin 2*70/9.8

**range**=793.63m

t= vsinθ/ g

=110 sin 70/9.8

time= 10.54 second

s=(110sin 70)*10.54 +1/2(-9.8)* (10.54^2)

s=1089-544.34

**height**=544.66m

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A 0.35-kg tennis racquet moving to the right at 20 m/s hits a 0.06-kg tennis ball that is moving to the left at 30 m/s. After the collision, the racquet continues to the right, but at a reduced speed of 10 m/s. What is the ball's velocity after the collision?

From conservation of momentum we have that:

[tex]0.35(20)-0.06(30)=0.35(10)+0.06v[/tex]Solving for v we have that:

[tex]\begin{gathered} 0.35(20)-0.06(30)=0.35(10)+0.06v \\ 7-1.8=3.5+0.06v \\ 0.06v=7-1.8-3.5 \\ 0.06v=1.7 \\ v=\frac{1.7}{0.06} \\ v=28.33 \end{gathered}[/tex]**Therefore the velocity of the ball after the collision is 28.33 m/s**

how long must a 50 N force act on a 400 kg mass to raise its speed from 10 m/s to 12 m/s

**Answer:**

see below

**Explanation:**

F=ma

50 = 400 a

A = 50/400 = .125 m/s^2

Accel = change in v / change in t

.125 = (12-10) / t Shows t = 16 seconds

Question 8 of 10 A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment? A. The brand of salt used B. The time it takes for the water to boil C. The kind of bottles used D. The amount of salt added to the water SUBMIT it's B

**Answer:**

**Explanation:**

A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment? A. The time it takes for the water to boil B. The amount of salt added to the water C. The kind of bottles used D. The brand of salt used

For friction on an incline, what direction does Fp point?

The **force **of **friction **on an inclined plane, will act **upward **to oppose the downward motion of the object on the inclined plane.

The **force **of **friction **is the force that opposes the motion of an object.

Since the **force **of **friction **opposes the motion of an object, it acts upward along the plane for an object moving along an inclined plane.

As the object moves downwards, the **force **of **friction **will act upwards to oppose the downward motion of the object.

Mathematically, the magnitude of **force **of **friction **on an object moving along inclined plane is given as;

Ff = μmg

where;

μ is the coefficient of frictionm is the mass of the objectg is acceleration due to gravityThus, the **force **of **friction **on an inclined plane, will act **upward **to oppose the downward motion of the object on the inclined plane.

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The complete question is below:

For friction on an incline, what direction does force of friction (Fp) point?

Circular Motion

The Trajectory of a mass point is a helix:[tex]r(t)=\left[\begin{array}{ccc}rcos(wt)\\rsin(wt)\\ht\end{array}\right]\\[/tex]

a) Compute the velocity v(t)=r(t) and the acceleration a(t).

b) What is the angel between the velocity and the acceleration?

c) If h=0, what is the radial acceleration and the velocity of the mass point at r=2m and 40rpm

No idea how to answer this question, I have never seen anything like it.

Any help would be awesome :)

**Answer and step by step explanation:**

First of all, I'm assuming you have had calculus, or this is going to be very awkward. Then, I'm replacing w with the Greek letter omega, it's a pet peeve of mine, sorry.

Shockingly, the derivative of a vector is computed by taking the derivative of each component (by linearity). If you've never heard the word derivative yet, you can think of the x and y components as two harmonic motions out of phase by 90°, and the z component as having constant speed h (harmonic motion is what you see when a mass moves along a circle with constant angular velocity and you look at it from the side).

At this point, let's use the derivative method for question a:

[tex]v(t)=\dot{r}(t)=\left[\begin{array}{ccc}-\omega r\sin(\omega t)\\\omega r\ cos(\omega t)\\h\end{array}\right]; a(t)=\dot v(t)=\ddot{r}(t)=\left[\begin{array}{ccc}-\omega ^2r\cos(\omega t)\\-\omega ^2r\ sin(\omega t)\\0\end{array}\right]\\[/tex]

Point b requires computing angles, which screams dot product to me.

[tex]\vec v(t) \cdot \vec a(t) =\left[\begin{array}{ccc}-\omega r\sin(\omega t)\\\omega r\ cos(\omega t)\\h\end{array}\right]\cdot \left[\begin{array}{ccc}-\omega ^2r\cos(\omega t)\\-\omega ^2r\ sin(\omega t)\\0\end{array}\right] = + \omega^3r^2 sin (\omega t) cos (\omega t) -\omega^3r^2 cos (\omega t) sin (\omega t) = 0 = ||\vec v|| ||\vec a|| cos \theta \implies cos\theta = 0 \implies \theta=\frac \pi2[/tex]

Now, the implied part is granted by the fact that we are assuming neither the velocity nor the acceleration are both zero, so the only option is for the cosine being zero, that makes the two vector orthogonal.

Finally, for point c, let's just take the moduli of both velocity and acceleration [tex]||\vec v|| = \omega r; ||\vec a||=\omega^2 r[/tex] and let's convert the angular velocity in civiliz... err, IS units: [tex]40 rpm = 40\times \frac{2\pi}{60}rad/s = \frac 43 rad/s[/tex]

Let's replace and we get

[tex]v=\frac83= 2.6 m/s\\a= \frac{32}3 = 10.5 m/s^2[/tex]

Imagine you are an alien from a distant galaxy. Your home planet does not have any gravity. You have just landed on Earth and make a few observations about the planet.

What are some ways you might first observe gravity?

How might you test gravitational force?

How would you describe gravity to other aliens back at your home planet?

Use details to support your answer.

The first thing an **alien **would notice about **gravity **on earth is how it affects their movement. If their planet is closer to the sun in their solar system, they'd be accustomed to a higher pull of gravity, hence weight. Based on the above, an alien will feel **lighter **on earth without auto-assistive **gravity adjuster technology**.

The simplest way to **test gravity **is by comparing my weight using "homemade" scales on the ship, then checking that against my weight off the ship.

The simplest way to describe Earth's gravity to those the home planet is to use scientific language. The acceleration of **gravity **at the Earth's surface is approximately **9.8** meters (32 feet) per second every second. This is constant.

Gravity, often known as gravitation, is a force that exists between all physical things in the universe. The force of **gravity **seeks to draw any two objects or particles with **nonzero mass **toward one other. Gravity affects things of all sizes, from subatomic particles to galaxy clusters.

Einstein proposed that the geometry of spacetime is what causes the force we call gravity. A mass (or energy) concentration, such as the Earth or sun, bends space around it in the same way as a boulder bends the flow of a river. As a result, **gravity **was created.

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A 2.55kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0360m . The spring has force constant 870N/m . The coefficient of kinetic friction between the floor and the block is 0.45 . The block and spring are released from rest and the block slides along the floor.

What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0160 m .)

The **speed **of the block when it has moved a **distance **of 0.0200 m from its initial position is 0.29 m/s.

The mass of the block is 2.55 Kg.** Coefficient of kinetic friction** is 0.45. **Spring constant** is 870N/m. The final compression of the spring is 0.0160 meters. The block is moved by distance of 0.0200 meters.

By considering the** block-spring system**, we can use Work energy theorem here,

According to **work energy theorem,**

Work done by **conservative forces** on the body + Work done by **none conservative forces** = Total change in** kinetic energy** of the body.

Work done by Conservative forces = 1/2Kx².

Where,

K is spring constant,

x is the **compression in spring,**

Work done by non-conservative forces = umgd

Where,

u is the coefficient of kinetic friction,

m is mass of the block,

d is the distance by which the block moves,

Here, we will take spring force as positive because the spring force and displacement both are in same direction,

Putting all the values,

1/2Kx² - umgd = 1/2mV² - 1/2mU²

1/2(870)(0.016)² - 0.45(2.55)(9.8)(0.02) = 1/2(2.55)V²

0.11136 - 0.22491= (1.275)V²

0.11355/1.275 = V²

V = 0.29 m/s.

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Set the cannon to have an initial speed of 15 m/s. For which situation do you think the cannon ball will go the highest: if it is set at a 25-degree angle, or if it is set at a 35-degree angle?

Question 3 options:

25 degrees

35 degrees

The cannon ball will travel the **highest distance **when the angle of projection is **35 degrees**.

The** maximum height** reached by a projectile is calculated using the following formula.

H = u²sin²θ/2g

where;

u is the initial velocity of the projectile θ is the angle of pojectiong is acceleration due to gravitywhen the angle of projection is 25 degrees;

H = (15² (sin 25)²) / (2 x 9.8)

H = 2.05 m

when the angle of projection is 35 degrees;

H = (15² (sin 35)²) / (2 x 9.8)

H = 3.78 m

Thus, the cannon ball will travel the **highest distance **when the angle of projection is **35 degrees**.

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Inside the nucleus, two protons are held together by a force which overcomes the repulsion. This force is called

Inside the nucleus, two** protons** are held together by a **strong nuclear** force which overcomes the repulsion. This is sometimes referred to as strong interaction.

The strong** nuclear force** or the strong interaction is strong enough to overcome the repulsive force between the two positively charged protons and thereby allowing protons and neutrons to stick together even in a small space. **Protons** are held together by the strong attractive nuclear force that binds together protons and neutrons.

The **strong force** is one of the four fundamental forces of nature that include gravity, electromagnetism, and the weak nuclear force. The strong force holds **protons** and neutrons within the nucleus of the atom and thus creating densest environments in nature.

The strong force dies off very fast with distance more than gravity or the **electromagnetic** force. It's difficult to detect the strong force outside of a nucleus.

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P and Q are points within a uniform electric field that are separated by a distance of 0.2 m as shown. The potential difference between P and Q is 75 V.How much work is required to move a +150 μC point charge from P to Q?0.023 J140 J0.011 J2800 J75 J

**ANSWER:**

0.011 J

**STEP-BY-STEP EXPLANATION:**

Given:

q = 150 μC

v = 75 V

r = 0.2

We calculate the work as the product between the charge and the potential difference, just like this:

[tex]\begin{gathered} W=q\cdot v \\ W=150\cdot10^{-6}\cdot75 \\ W=0.01125\text{ J} \\ W\cong0.011\text{ J} \end{gathered}[/tex]The work required is 0.011 J

PLS help due today number in the picture are 15 and 0

As the horizontal **component **of the velocity is constant throughout the projectile, the time of the flight of the **projectile **is independent of the **horizontal **velocity, therefore the correct answer is option E .

It is the **motion **of any object or body when it is ejected off the surface of the earth and follows any curving path while being affected by the **gravitational **pull of the planet .

The total **time **of the flight of the **projectile **= 2uSinθ / g

Thus , the time of the flight of the **projectile **is independent of the **horizontal **velocity, therefore the correct answer is option E.

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Giant electric eels can deliver a shock of 590 V with up to 1.0 A of current for a brief time. A snorkeler in salt water has a body resistance of about 770 Ω. A current of about 500 mA can cause heart fibrillation and death if it lasts too long.

A)What is the maximum power max a giant electric eel can deliver to its prey?

B)If the snorkeler is struck by the eel, what current will pass through her body?

C)Is this current large enough to be dangerous?

D)What power received does the snorkeler receive from the eel?

**Answer:**

**Explanation:**

Given:

U = 590 V

I = 1.0 A

R₁ = 770 Ω

I* = 500 mA = 0.5 A

_________________

P - ?

A)

**Power**:

P = U·I = 590·1.0 = 590 W

B)

**Сurrent**:

I₁ = U / R ₁ = 590 / 770 = 0.770 A = 770 mA

C)

**This current is life-threatening!!!**

I₁ > I* (700 mA > 500 mA)

D)

**Energy received:**

E = I₁²·R₁ = 0.770²· 770 ≈ 460 W

What is the electric field between the plates of a capacitor that has a charge of 10.75 microC and voltage difference between the plates of 97.87 Volts if the plates are separated by 2.11 mm?

The eletric field on a capacitor is given by the following formula:

[tex]E=\frac{Q}{A\epsilon_0}=\frac{V}{d}[/tex]Where Q is the difference of charge between plates, A is the area, and e0 is a constant. It can also be written as the second formula, where V is the voltage, and d is the distance between plates.

When we use the second formula, with the values from the exercise, we get:

[tex]E=\frac{97.87}{2.11*10^{-3}}=46383.886[\frac{V}{m}][/tex]**Thus, our final answer is E=46383.886 V/m**

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